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Mathematics MasterclassesStretching the imagination

Editedby

MICHAEL SEWELL

Professor of Applied Mathematics, University of Reading

OXFORD NEW YORK TOKYO

OXFORD UNIVERSITY PRESS

1997

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Mathematics Masterclasses-Stretching the imagination / edited by Michael Sewell.1. Mathematics 1. Sewell, M. J. (Michael1.), 1934-

QA3.M3631997 51O-dc20 96-27450

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Midsomer Norton, Avon

6 Discrete Mathematics and itsApplication to Ecology

by PHILIP MAINICentre for Mathematical Biology, Mathematical Institute, Universityof Oxford

1. Introduction

In this chapter we will illustrate how simple mathematical ideas can be used tounderstand how a population grows. We first consider the well-known problemwhich generates the Fibonacci sequence. Consider a population of rabbits.Suppose that every pair of rabbits can reproduce only twice, when they are one-and two-months old, and that each time they produce exactly one new pair ofrabbits. Assume that all newborn rabbits survive. The problem here is to see ifwe can predict how many newborn rabbits there will be in a certain generationgiven an initial newborn rabbit population.

Before beginning, we need to use a convenient notation. Let us define thequantity Nm to be the number of newborn pairs in generation m. Suppose thatwe begin with one newborn pair. Then we say that No, the number of newbornpairs in the zero generation (the Adam and Eve of rabbits), is one. Now, afterone month, this pair of rabbits gives birth to one pair of rabbits. Therefore, wehave that NI, the number of newborn pairs in the first generation, is one. At theend of the second month, Adam and Eve give birth to their second newborn pairand then will no longer give birth to any more rabbits. But, the pair of rabbits inthe first generation also give birth to a pair of newborns (Adam and Eve's'grandbunnies'). Therefore, N2, the number of newborn pairs in the secondgeneration, will be 1 + 1 = 2. In the third generation, newborn pairs will arisefrom those pairs that were newborns in the second generation (N2), and fromthose newborns in the first generation (NI), Therefore the number of newbornsin the third generation, N3, will be N2 + NI = 2 + 1 = 3. The number in thefourth generation, N4, will be N3 + N2 = 3 + 2 = 5 (see Fig. 1). We can nowprove the following theorem.

Theorem 1 (The Rabbit Problem)If

Nm = number of newborn pairs in generation m

then

Nm+l =Nm +Nm-l' (1)

uu Ulscrete ivlaltlt::1ilal1csand Hs Application to EcologyDiscrete Mathemal1cs aud Its Apphcal10n to Ecology In

ProofThe number of newborn rabbits in the (m + nth generation will come eitherfrom one-month old parents who were born in the mth generation or fromtwo-month old parents who were born in the (m - nth generation (see Fig. 1).Hence result. 0

To see how this ties in with the above discussion, let us suppose that No = 1(this means that we begin with one pair of newborns in the zero generation).Then, we know that NI = 1. Now, if we put m = 1 into equation (1), we havethat N2 = NI + No, that is, Nz = 1 + 1 = 2, which is the number of newborns inthe second generation. Now putting m = 2 into equation (1), we have thatN3 = N2 + NI = 2 + 1 = 3, the number of newborns in the third generation. Form = 3, we have N4 = 3 + 2 = 5, the number of newborns in the fourth genera-tion. Continuing like this we have that N5 = 5 + 3 = 8, N6 = 8 + 5 = 13, N7 =13 + 8 = 21, etc.

The sequence of numbers 1,1,2,3,5,8, 13,21, . .. is known as the Fibonaccisequence. Equation (1) is an example of a recu"ence relation and this type ofrelation is very common in ecological studies. A mathematical formulation ofsuch a problem is known as a mathematical model of the problem. The numbersgenerated by the Fibonacci sequence are known as the Fibonacci numbers andoccur widely in nature. For example, the numbers of spirals running in oppositedirections on a ripening sunflower or a pine cone are often consecutiveFibonacci numbers.

We now move on to what appears, to begin with, to be a very differentproblem. We call it the Beetle Problem. Suppose that beetles of one generationproduce the larvae (immature form of an insect which eventually becomes theinsect) of the next generation and then die so that only the larvae survive.Assume that all the larvae survive. The problem here is, can we predict thenumber of larvae in a certain generation? In this case, we define Nm to be thenumber (in some units) of larvae in generation m. Suppose that the average

number of larvae produced per beetle is 2. Let us assume that in generation 0,No is 1. It is important to note here that setting NI to be one does notnecessarily mean that in the first generation there is one beetle. This is becausewe have not specified our units. For example, if the unit was thousands, thenNo = 1 would mean that there were one thousand beetles in generation O. Notethat No = ! would not mean half a beetle, but would mean 500 (one half of athousand) beetles.

Assuming that No = 1, we have that NI, the number of beetles in the firstgeneration, is 2 X 1 = 2. The beetles in the second generation arise from thelarvae produced by the first generation, so Nz is 2 X NI = 4. In the same way,N3 = 2 X 4 = 8 etc. In fact, we can now prove the following theorem.

Theorem 2 (The Beetle Problem)If

Nm = number of beetles (in some units) in generation m

r = average number of larvae produced per beetle

then

Nm = Norm (2)

w~W~W~

ProofThe generation m beetles come from the (m - nth generation. Now eachbeetle in the (m -nth generation produced r larvae, so Nm= rNm_I'Applyingthe same argument to the (m - nth generation, we have Nm-I = rNm_z. So wecan write Nm= rZNm_z'If we carry on this reasoning we deduce that Nm= Norm~~n 0

For our example, r = 2 and No = 1. The formula then says that NI = No X 2 =2, Nz = No X 4 = 4, N3 = No X 8 = 8, etc.

Notice that if we want to calculate the number of beetles in a certain

generation, then all we need to know is No and r and we have immediately thatNm = Norm. However, for the rabbit problem, to find Nm, we need to knowNm-I and Nm-z. But to calculate these values, we need to know Nm-3' etc.Therefore, if m is very large, then the calculations involved for the beetleproblem are easier than those for the rabbit problem. However, in question 6 ofWorksheet 1 we will see that there is an easier way to calculate Nm for therabbit problem, when m is large.

1. Consider the Fibonacci numbers 1,1,2,3,5,8,13,21,34,55,89,144,233,...(a) Calculate (to 3 decimal places) the fractions obtained from dividing each

b b h.

b h . 1 Z 3 5 8 13 21 34 55 89 144 233num er y t e prevIous num er, t at IS, T, T, 2' 3' 5' g, 13' 21' 34"'55' 89' 144'(b) What do you notice about the values you calculated in (a)?(c) The calculations in (a) and (b) suggest that for two consecutive numbers

N

Nm+ I' Nm, in the Fibonacci sequence, ~::= r for large m. What is thatNm

value of r?

IW~

I IW~

11 &i~W~ .I

w~

I

W~I

w~W~

w~Fig. 1. Proof of Theorem 1 (see text for details)

92Discrete Mathematics and its Application to Ecology Discrete Mathematics and its Application to Ecology 93

, -,~'-L ' .- I

I r-

--------

5. Calculate Np Nz, N3, N4 and Ns (each to 2 decimal places) for equation (3)in the case where r = 3.1, No = 0.70. What do you think are the values ofN6, N7, Ns, Ng,...?

6. In question 1, we made an approximate calculation for r. We can make aNm+ I Nm

more exact calculation as follows. Suppose that - = - = r.Nm Nm - I

(a) From equation (1) show t;at Nm+ I = ( 1 + ~ )Nm.(b) From (a), show that 1 + - = r.

r(c) From (b), calculate r.

-I

2. The logistic map

In Worksheet 1, question 3, we saw that the models presented in Section 1 havetheir limitations, and we introduced a new model

Nm+ 1= rNm(1.0 - Nm).

This model is called the logistic model, or the logistic map. Before we analysethis model further we need the following definition.

(4)

Fig. 2. The Beetle problem for the case r = 2. Note the differences between this figureand Fig. 1

(d) Is the Rabbit problem really that different from the Beetle problem?2. From equation (2) calculate NI, Nz, N3 and N4 in the following cases.

(a) No = 2.0, r= 2.0.(b) No = 2.0, r = 0.5.

3. From equation (2) what size will Nm be for m large in the following cases.(a) r > 1.0.(b) r = 1.0.(c) r < 1.0.

(d) Do you think that equations (1) and (2) are good mathematical models?

For the following exercises assume that the recurrence relation has the form

Nm+1= rNm(1.0 - Nm). (3)

(Note that if Nm is small, equation (3) looks a bit like equation (2».

4. Calculate NI, Nz, N3 and N4 in the following cases.(a) No= 0.6, r = 0.5.(b) No = 0.4, r = 2.0.(c) No = 0.6, r = 2.0.(d) No = 0.5, r = 2.0.

(e) In (b), (c) and (d) what do you think the value of Nm would be for m verylarge?

(f) Is this a better mathematical model than equations (1) and (2)?

DefinitionIf the successive values calculated from a recurrence relation tend towards somefixed value, we call this value the steady state value. It has the property thatNm+ 1 = Nm = N*, the steady state value, for large m.

The steady states for equation (4) are given byr-1

N* = 0 or N* = -.r

(5)

For 0 < r < 1 the second steady state does not make sense because it is negativeand if we start off with 0 ~ No ~ 1 in equation (4) then 0 ~ Nm ~ 1 for every m.So the only sensible solution is the first solution of equation (5) (compare thiswith Worksheet 1, question 4(a». For 1 < r < 3 the second solution makes senseand it appears from Worksheet 1, question 4(e), that this is the steady state, atleast for r = 2.0. Hence, although there are now two possibilities for the steadystate, we have moved off the first solution onto the second. This is called abifurcation. For r = 3.1, the second solution of equation (5) is still a sensiblesteady state but Works heet 1, question 5 suggests that we have a 2-cycle solution.We may summarize these results on the bifurcation diagram in Fig. 3. Note thatwe have been careful not to go too far beyond r = 3 in case there are yet moresurprises in store!

We can get an idea of what the solution to equation (4) will look like by usinga graphical technique called cobwebbing. The technique is illustrated in Fig. 4and the results represented graphically in Fig. 5.

Discrete Mathematics and its Application to Ecology

1.0

Discrete Mathematics and its Application to Ecology

1.0

95

N* Nrn

0.8

0.6

0.4

0.2

0.00.5 1.0 1.5 2.0 2.5

r3.0 3.5 4.0

Fig. 3. Bifurcation diagram for equation (4) showing how the steady state N* varieswith r. The dashed lines are unstable solutions

0.8

0.6

0.4

8 10 12 14

0.2

0.02 64

m

Fig. 5. Graphical representation of the solution obtained by the cobwebbing techniqueillustrated in Fig. 4 (Compare this solution with'that from Worksheet 1, question 4)

We draw the curve of equation (4) (with r = 0.5) and also the straight lineNm+ I = Nm.We first find the value 0.6 along the horizontal axis.This is No.NowNI = 0.5 X NoG - No)' But this is exactly the length of the vertical straight linefrom the point No to the curve. If we then draw from there a line parallel to thehorizontal axis, then the distance this line is above the horizontal axis is always

NI, If we drop a perpendicular line onto the horizontal axis from the pointwhere this line intersects the Nm+I = Nm line, we have that the distance fromthis point to the origin is also NI, That is, given No, we have been able tocalculate, graphically, the value of NI, In a similar way we can now go on tocalculate N2, N3,..., etc.1.0

Nm+1

0.8

0.6

0.4

0.2

0.00.2 0.4 0.6

No0.8 1.0

N2 NI N m

Fig. 4. Illustration of the cobwebbing technique for equation (4) with No = 0.6, r = 0.5

N =Nm+J m

0.0 0.6 0.80.2 0.4Nm

Fig. 6. Cobweb for Worksheet 2, question 1

1.0

Nm+1

0.8

0.6

0.4

0.2

96 Discrete Mathematics and its Application to Ecology

Worksheet 2

1. Use Fig. 6 to cobweb the solution to equation (4) for the case No = 0.7,r= 2.0.

2. Illustrate your results to question 1 graphically. Does this solution cycle ordoes it go to a steady state?

3. Use your calculator to find Nj, Nz, N3, N4, Ns, N6, N7, and N8 (each to 3decimal places) for equation (4) in the case No = 0.501, r = 3.5. Draw thecobweb diagram on the following figure (Fig. 7).

1.0

Nm+l

0.8

0.6

0.4

0.2

Nm+l=3.5Nm(1.0-Nm)

0.0 0.2 0.4 0.6 0.8 1.0N

III

Fig. 7. Cobweb for Worksheet 2, question 3

4. Illustrate your results to question 3 graphically.5. From equation (4), what is the equation for Nm+z in terms of Nm? What are

the steady states of this equation?

3. Chaos

From Worksheet 2, question 3, we saw that if r = 3.5, then equation (4) has a4-cycle solution. As r increases further, we get an 8-cycle solution and then a16-cycle solution, etc. We can represent these results (Fig. 8) on the bifurcationdiagram we began to draw in Fig. 3.

If we increase r beyond a critical value re :::::3.83 we lose the periodic structureof the solution. Figure 9 illustrates the case when r = 3.9. In this case thesolution does not have a periodic structure. It is said to be chaotic. If we were tochange the value of No slightly we would get a completely different lookingsolution. Another chapter in this book, 'A Little Bit of Chaos', introduces you toother aspects of chaos.

Discrete Mathematics and its Application to Ecology

1.0

N*

0.8

0.6

0.4

0.2

0.0 0.5

97

4.01.0 1.5 2.5 3.0 3.52.0r

Fig. 8. The completed bifurcation diagram illustrating, schematically, the cascade ofperiod doubling bifurcations

Equation (4) is a good model for insect growth - many insect populationseither reach a steady state or oscillate in a 2-cycle (see Fig. 10).

Nm

1.0

0.8

0.6

5 10 15 20 25

0.4

0.2

m

Fig. 9. The solution to equation (4) for the case r = 3.9, No = 0.2. Note that for somegenerations the population gets very large then crashes back in the following generation

98 Discrete Mathematics and its Application to Ecology

1.5

1.0 1.0

0.5

I I I I " I I

V\.A-~0.5

0.0

(a)

0.0

(b)

105 5 10

Fig. 10. Examples of population growth of two types of beetles. The vertical axis is the

number of beetles(thousands),the horizontalaxisisthe generation number. (Redrawnfrom Hassel (1976))

Note that for r> 4 the model predicts negative populations and thus is notrealistic in this case. You may want to do a few iterations or draw some cobwebmaps to see why (hint: start off with r = 4.1, No = 0.5).

The analysis of the model equation (4) suggests that if we let r get too largethen we will get cycling populations with very large numbers or chaotic popula-tions with the possibility of very large numbers. Therefore, in order to controlthe population numbers, we really need to keep r < 3.83 and preferably muchsmaller. We can decrease r by, for example, releasing sterile insects into thepopulation, so that the average birth rate per insect decreases.

We can write down more complicated relations, or iterated maps. For example,we could have

rNmb

Nm+ 1 = 0.0 + aNm)(6)

where r, a and b are positive constants. Figure 11 shows the results fromequation (6) for the case r = 20, a = 0.01, No = 0.01 and different values of b.

4. Summary

In this chapter we have illustrated how mathematical models can be used tostudy population growth. The analysis of these models determines if the assump-tions we made in forming the model are sensible. If not, we have to reformulatethe model. This helps us to understand more deeply the processes of populationgrowth that we are trying to model and also predicts how the population willreact if we carry out certain experiments. The role of a predictive mathematicalmodel is very important. For example, it can aid our understanding of howcertain diseases spread within a population and help us to choose the bestcontrol strategy. Mathematical models are important not only in ecology, butalso in the fields of biology, physics, chemistry and medicine.

Discrete Mathematics and its Application to Ecology99

40 ,.-.- 4

330

';?,E<"'>0- 20

';i?"E"b- 2

10

15m

20 250100 ._-e-.A I

2 4 6 8 10 12 14m

5

(b)(a)

20

15

E';i?";:; 10

5

0 I I I I

5 10 15 20 25m

(c)

Fig. 11. The solution to equation (6) for the case r = 20, a = 0.01, No = 0.01 anddifferent values of b: (a) b = 0.5, (b) b = 2.5, (c) b = 5.0. The solution shown in (a) is verysimilar to the behaviour of growing bacteria and of yeast

Solutions to Worksheet 1

1. (a) 1, 2, 1.500, 1.667, 1.600, 1.625, 1.615, 1.619, 1.618, 1.618, 1.618, 1.618.(b) The values calculated in (a) tend to 1.618 to 3 decimal places. (In fact,

1 + /5this is the golden mean ).2Nm+l

(c) -::::: r, where r:::::1.618to 3 decimal places.Nm

(d) The Rabbit problem, for large rn, is a special case of the Beetle problemwith r = 1.618 to 3 decimal places.

2. (a) 4.0, 8.0, 16.0, 32.0.(b) 1.0, 0.5, 0.25, 0.125.

3. (a) Nm will be very large.(b) N", = No.

100 Discrete Mathematics and its Application to Ecology

(c) Nm will be very small.(d) Equation (1) predicts that the population of rabbits will grow forever;equation (2) predicts that the population of beetles will either grow and grow(if r> 1.0), or die down to zero (if r < 1.0) or stay at its initial value (ifr = 1.0). Neither of these models is very good because populations do notbehave in this way. For example, death has been ignored.

4. (a) 0.120, 0.053, 0.025, 0.012 to 3 decimal places.(b) 0.480, 0.499, 0.500, 0.500 to 3 decimal places.(c) 0.480, 0.499, 0.500, 0.500 to 3 decimal places.(d) 0.500, 0.500, 0.500, 0.500.(e) Nm tends to 0.5 for large m.(f) This seems a better model because for certain r-values the populationreaches a steady value.

5. 0.65, 0.71, 0.65, 0.71, 0.65, 0.71, 0.65, 0.71,. ..6. (a) From the assumption, Nm-I = Nmlr, therefore equation (1) becomes

Nm+ I = Nm +Nmlr.(b) From (a), we have Nm+1= (1 + 1/r)Nm. The assumption says thatNm+I = rNm. Therefore 1 + 1/r = r.(c) From (b), rZ - r - 1 = O. Therefore 2r = 1 :i: {5. The negative solutiondoes not make sense because r is assumed positive, therefore r = (1 + {5)/2.

Solutions to Worksheet 2

1.

1.0

0.8

Nm+l

0.6

0.4 t /1: ___m,Y--~r ,'" ,'" ,'" ,'" ,'" ,'"

t' ", ", ", ", ", ", ", ", ", "

0.2

0.00.2 0.4 \ 0.6 0.8 1.0

NI NZ N3 NO Nm

Fig.12 Cobweb diagram for Worksheet 2, question 1

Discrete Mathematics and its Application to Ecology 101

2.1.0

~

Nm

0.8

.0.6

./-.-.---.-.-.--.---0.4

0.2

2 4 6 8 10 12 14m

Fig. 13 Graphical representation of the solution obtained by the cobwebbing tech-nique for Worksheet 2, question 1. The solution goes to a steady state

3. 0.875, 0.383, 0.827, 0.501, 0.875, 0.383, 0.827, 0.501

1.0

0.8

Nm+!

0.6

0.4 V, l :\t

'T "

: um_~m--m_--~u i" ," ," ," ," ," ," ," ," ," ," ," ,, , .,

0.2

,,: \N +J=3.5N (I.O-N ),m m m,,,,

0.00.2 0.4 0.6

NZ NO Nm

0.8

N3 NI

Fig 14 Cobweb diagram for Worksheet 2, question 3

lLJ:L Discrete Mathematics and its Application to Ecology

4. 1.0

Nm

0.8

0.6

0.4

0.2

2 4 6 8 10 12 14m

Fig 15 Graphical representation of the solution obtained by the cobwebbing tech-nique for Worksheet 2, question 3

5. From equation (4) we can write Nm+2 = rNm+10.0 - Nm+ 1)' Using equation (3) again,we have Nm+1= rNmO.O- Nm),so we can write

Nm+2= r2NmO.0 - N",)(1.0 - rNm(1.0 - Nm)).

The steady states of equation (7) are given by Nm+2= Nm = N*. Thus we have

N* = r2N*(1.0 - N*)(1.0 - rN*(I.O - N*))which has solution N* = 0 or

(7)

(8)

(

1

)

1 1(N*)3 - 2(N*)2 + 1 + - N* + - - - = O.

r r3 r

1We know that 1 - - is also a steady state so we can factor this out from equation (9)rto get

(9)

(N*-1+~)((N*)2-(1+~)N*+~+ :2)=0. (10)

Thus the two further solutions are given by the roots of ((N*)2 - 0 +l/r)N* + 1/r - 1/r3) = 0, that is, 2N* = 1 + 1/r:t [0 + 1/r)2 - (4Ir)(1 + 1/r)]I/2.This may be written as 2N* = 1 + (l/r):t O/r)[(r - 1)2 - 4]1/2. Note that this solu-tion will make sense only if r> 3, so we have a 2-cycle solution possible if r> 3.

References

R. Anderson and R. M. May, 'The logic of vaccination,' New Scientist (18th November,1982), 410-15

O. M. Hassel, The dynamics of competition and predation (Edward Arnold Publishers,1976)

R. M. May (editor), Theoretical Ecology, Principles and Applications (2nd edition) (Black-well Scientific Publications, 1981)